Re: MySQL and PHP question

2002-12-18 Thread C. Reeve
Thanks for the info. I tried using mysql_free_result()  and mysql_close($db)
and opening the DB again for the second query with the same results. I
haven't tried using a different variable for $result.

- Original Message -
From: "Steve Yates" <[EMAIL PROTECTED]>
To: "MySQL List" <[EMAIL PROTECTED]>
Sent: Wednesday, December 18, 2002 1:32 AM
Subject: Re: MySQL and PHP question


> On Wed, 18 Dec 2002 22:54:53 -0500, C. Reeve wrote:
> >while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
> >$num1 = $line['number'];
> >}
>
> Loop through the result and return the value from the last row?
>  Did you want "select sum(number) as SubTotal1 from table1 where
> userid=$user1"?
>
> >$query "select number from table2 where userid=$user2";
> >$result = mysql_query($query) or die("Query failed");
>
> You re-used the same variable $result.  Offhand I would try
> using a second variable, and/or using mysql_free_result() in between.
>
>  - Steve Yates
>  - My hard disk is full!  Maybe I'll try this message section thing.
>
> ~ Taglines by Taglinator - www.srtware.com ~
>
>
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Re: MySQL and PHP question

2002-12-17 Thread Steve Yates
On Wed, 18 Dec 2002 22:54:53 -0500, C. Reeve wrote:
>while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
>$num1 = $line['number'];
>}

Loop through the result and return the value from the last row?
 Did you want "select sum(number) as SubTotal1 from table1 where
userid=$user1"?

>$query "select number from table2 where userid=$user2";
>$result = mysql_query($query) or die("Query failed");

You re-used the same variable $result.  Offhand I would try
using a second variable, and/or using mysql_free_result() in between.

 - Steve Yates
 - My hard disk is full!  Maybe I'll try this message section thing.

~ Taglines by Taglinator - www.srtware.com ~


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Re: MySQL and PHP question

2002-12-17 Thread C. Reeve
 I've since deleted the code and did it another way. The code however went
something like this. This is just a mock up (there is more being selected
etc and there are some joins, but the syntax is exact).

So what happens is the select statement with $user1 in it gets the proper
values (the first one in the example), the other one doesn't, but if I
change $user1 and $user2 around the second select will get the values.

So on the surface it sounds like a problem with $user2, but I have run
extensive tests and $user2 gets the proper values (they both come from a
form).

In another project I am doing, I tried a similar feat with totally different
values and got the same results. This time I was trying to do one SELECT and
one UPDATE.

If anyone has any ideas it would be appreciated. Also, if there is a better
way to do the following this would also be appreciated.

$query "select number from table1 where userid=$user1";
$result = mysql_query($query) or die("Query failed");

while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
$num1 = $line['number'];
}

$query "select number from table2 where userid=$user2";
$result = mysql_query($query) or die("Query failed");

while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
$num2 = $line['number'];
}

$total = $num1 + $num2;

mysql_query("UPDATE table3 SET total = '$total' WHERE userid=$user");

TIA

- Original Message -
From: "Steve Yates" <[EMAIL PROTECTED]>
To: "MySQL List" <[EMAIL PROTECTED]>
Sent: Tuesday, December 17, 2002 5:12 PM
Subject: Re: MySQL and PHP question


> On Wed, 18 Dec 2002 16:05:35 -0500, C. Reeve wrote:
>
> >I can not use two mysql queries in the same PHP file
>
> Please post code snippets, this is possible.
>
>  - Steve Yates
>  - Burger Borg:  We do it our way; yours is irrelevant.
>
> ~ Taglines by Taglinator - www.srtware.com ~
>
>
> -
> Before posting, please check:
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<[EMAIL PROTECTED]>
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RE: MySQL and PHP question

2002-12-17 Thread Fraser Stuart
I can...

Hard to know why without more info but a starting point would be to turn on
the query log in MySQL and see what queries it is receiving.

_

 Fraser Stuart
 Logistics IT

 77-85Phone: +61 2 9335 1235
 Roberts Rd  Mobile: +61 419 233 732
 Greenacre NSW [EMAIL PROTECTED]
 Australia 2190  www.toll.com.au
_

| -Original Message-
| From: C. Reeve [mailto:[EMAIL PROTECTED]]
| Sent: Thursday, 19 December 2002 8:06 AM
| To: MySQL List
| Subject: MySQL and PHP question
|
|
| Hi,
|
| Why is it that I can not use two mysql queries in the same PHP
| file. I have
| tried to do this several times trying to put my site together and have had
| nothing but problems.
|
| The recent one is this:
|
| I use a SELECT statement to get some info from a table, then I modify that
| info and use an UPDATE statement to update another table. The update does
| not work (all variables used by PHP contain the proper values as does the
| info selected from the table).
|
| I get no errors as it appears the update worked, but the table was not
| updated.
|
| I have also found that I can not do two select statements in the same PHP
| file, as the second one will not contain any data . The query
| does not fail,
| there is just no data.
|
| Any info is appreciated.
|
| TIA
|
|
|
|
|
| -
| Before posting, please check:
|http://www.mysql.com/manual.php   (the manual)
|http://lists.mysql.com/   (the list archive)
|
| To request this thread, e-mail <[EMAIL PROTECTED]>
| To unsubscribe, e-mail
| <[EMAIL PROTECTED]>
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|
|


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Re: MySQL and PHP question

2002-12-17 Thread Steve Yates
On Wed, 18 Dec 2002 16:05:35 -0500, C. Reeve wrote:

>I can not use two mysql queries in the same PHP file

Please post code snippets, this is possible.

 - Steve Yates
 - Burger Borg:  We do it our way; yours is irrelevant.

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Re: MySQL and PHP question

2002-12-17 Thread Chris Boget
> Why is it that I can not use two mysql queries in the same PHP file. 

You can, I assure you.

> Any info is appreciated.

Perhaps if you could provide your code?  W/o it there is no way we can
determine where the problem might be coming from.

Chris


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Re: MySQL and PHP Question

2001-02-08 Thread Ing . Alejandro Vázquez C .

Adrian D'Costa wrote:

> Hi,
>
> I have two tables city and country
> +---+-+--+-+-++
> | Field | Type| Null | Key | Default | Extra  |
> +---+-+--+-+-++
> | id| int(5)  |  | PRI | 0   | auto_increment |
> | city  | varchar(35) |  | MUL | ||
> | countryid | int(5)  |  | MUL | 0   ||
> +---+-+--+-+-++
> 3 rows in set (0.00 sec)
>
> mysql> desc country;
> +-+-+--+-+-++
> | Field   | Type| Null | Key | Default | Extra  |
> +-+-+--+-+-++
> | id  | int(5)  |  | PRI | 0   | auto_increment |
> | country | varchar(35) |  | MUL | ||
> +-+-+--+-+-++
> 2 rows in set (0.00 sec)
>
> I am trying to write a php script that does the following.
> In an html page I have two dropdowns one country which will get the data
> from the country table and city which should get the data from the city
> table based on the one selected from the country drop down.  I do not want
> to use a submit button.  I use a javascript and using onChange pass the
> country.id to the javascript but do not know how to pass it back to the
> php script.  Any pointers
>

You have to submit via javascript the country id whenever the dropdown selection
changes. You have to be carefull using a hidden field so your php-script
recognices when you pushed your form submition button or you just had changed
your country.



>
> Adrian
>
> --
> ===
> Adrian D'Costa
> [EMAIL PROTECTED]
> www.pcsadvt.com
> ===
>
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RE: MySQL and PHP Question

2001-02-08 Thread Cal Evans

This is really a JavaScript problem if you don't want to use the submit
button.

Use PHP to load the values of both tables into a JavaScript array.  Then
write JS code that will change the contents of the City control based on
what is selected in the country control.  Not terribly difficult.

Cal
http://www.calevans.com


-Original Message-
From: Adrian D'Costa [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 06, 2001 12:01 AM
To: [EMAIL PROTECTED]
Subject: MySQL and PHP Question


Hi,

I have two tables city and country
+---+-+--+-+-++
| Field | Type| Null | Key | Default | Extra  |
+---+-+--+-+-++
| id| int(5)  |  | PRI | 0   | auto_increment |
| city  | varchar(35) |  | MUL | ||
| countryid | int(5)  |  | MUL | 0   ||
+---+-+--+-+-++
3 rows in set (0.00 sec)

mysql> desc country;
+-+-+--+-+-++
| Field   | Type| Null | Key | Default | Extra  |
+-+-+--+-+-++
| id  | int(5)  |  | PRI | 0   | auto_increment |
| country | varchar(35) |  | MUL | ||
+-+-+--+-+-++
2 rows in set (0.00 sec)

I am trying to write a php script that does the following.
In an html page I have two dropdowns one country which will get the data
from the country table and city which should get the data from the city
table based on the one selected from the country drop down.  I do not want
to use a submit button.  I use a javascript and using onChange pass the
country.id to the javascript but do not know how to pass it back to the
php script.  Any pointers

Adrian

--
===
Adrian D'Costa
[EMAIL PROTECTED]
www.pcsadvt.com
===


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